Question: A particle moves along the curve $\dfrac{x^2}{y^2}=25$ so that that the $y$ -coordinate is increasing at a constant rate of $3$ units per minute. What is the magnitude (in units per minute) of the particle's velocity vector when the particle is at the point $(-5,1)$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $-5$ (Choice B) B $-2\sqrt{3}$ (Choice C) C $\sqrt{34}$ (Choice D) D $3\sqrt{26}$
Background When working with motion along a curve, we should remember that this motion can also be represented by a position vector $(x,y)$. The main difference is that we're not given the equations for $x$ and $y$ in terms of time $t$, but only the relationship between $x$ and $y$ themselves. This however shouldn't prevent us from assuming the expressions for $x$ and $y$ in terms of $t$ exist. As always, $\vec{v}(t)=\left(\dfrac{dx}{dt},\dfrac{dy}{dt}\right)$. Setting up the math We are given that $\dfrac{dy}{dt}=3$ for any value of $t$. We are asked for the magnitude of the particle's velocity vector when the particle is at the point $(-5,1)$. In other words, we are asked for $\left|\left|\left(\dfrac{dx}{dt},\dfrac{dy}{dt}\right)\right|\right|$ at the point $(-5,1)$. Finding $\dfrac{dx}{dt}$ $\dfrac{dx}{dt}=\dfrac{3x}{y}$ Finding $\dfrac{dx}{dt}$ at $(-5,1)$ The expression for $\dfrac{dx}{dt}$ depends on both the particle's $x$ -coordinate ${-5}$ and its $y$ -coordinate ${1}$ : $\begin{aligned} \dfrac{dx}{dt}&=\dfrac{3({-5})}{({1})} \\\\ &=-15 \end{aligned}$ Therefore, the particle's velocity vector at the point $(-5,1)$ is $(-15,3)$. Finding $||(-15,3)||$ $||(-15,3)||=3\sqrt{26}$ In conclusion, the magnitude of the particle's velocity vector when the particle is at the point $(-5,1)$ is $3\sqrt{26}$ units per minute.